题解 | #牛牛的单链表求和#
牛牛的单链表求和
https://www.nowcoder.com/practice/a674715b6b3845aca0d5009bc27380b5
#include <stdio.h>
#include <stdlib.h>
//声明结构体
struct List
{
int date;
struct List* next;
};
//新节点
struct List* BuyNewCode(int x)
{
struct List* new=(struct List*)malloc(sizeof(struct List));
new->date=x;
new->next=NULL;
return new;
}
//尾插（不带头）
void PushBack(struct List** pplist,int x)
{
struct List* new=BuyNewCode(x);
if(*pplist==NULL)
{
*pplist=new;
}
else
{
//用于寻找尾节点的指针
struct List* tail=*pplist;
while(tail->next!=NULL)
{
tail=tail->next;
}
//找到，链接新节点
tail->next=new;
}
}
//求和
int Sum(struct List* plist)
{
struct List* cur=plist;
int sum=0;
while(cur!=NULL)
{
sum+=cur->date;
cur=cur->next;
}
return sum;
}
int main()
{
int n=0;
scanf("%d",&n);
//不带头指针的链表
struct List* plist=(struct List*)malloc(sizeof(struct List));
plist=NULL;
int i=0;
int k=0;
for(i=0;i<n;i++)
{
scanf("%d",&k);
PushBack(&plist,k);
}
int ret=Sum(plist);
printf("%d\n",ret);
return 0;
}